2
$\begingroup$

English is not my first language, so I'm sorry if I'm not very clear. I can clarify any question you have. Also, I don't know how to use that math formatting so I apologize for it.

I was asked to come up with a formula from the following image: image

And the "hint" was: $1 + 2\times2^2 + 3\times3^2 + 4\times4^2+ 5\times5^2= \cdots$

It seems like a way to calculate the area of a square:

$$k^2 = 1 + 2\times2^2 + \cdots + l \times l^2$$

Where $1 + 2 + \cdots + l = k$, and $k$ is the side of the square

I tested this with some numbers:

$$\begin{align} 10^2 &= 1 + 2 \times 2^2 + 3 \times 3^2 + 4 \times 4^2\\ 100 &= 1 + 8 + 27 + 64\\ 100 &= 100\\ \end{align}$$

I have to prove this using induction so I tried this:

$$(k+n)^2 = 1 + 2 \times 2^2 + \cdots + l \times l^2 + n \times n^2$$

If I understood induction correctly, I plug the first formula inside the second one:

$$(k+n)^2 = k^2 + n \times n^2$$

…and this is where I'm stuck. I can't figure out how to make these two equal, even though if I substitute them with numbers it appears the formula is correct.

$\endgroup$
6
  • $\begingroup$ $$\sum_{k=1}^nk*k^2=\frac{n^2(n+1)^2}4$$ $\endgroup$
    – RE60K
    Sep 17 '14 at 19:05
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$
    – Alice Ryhl
    Sep 17 '14 at 19:08
  • $\begingroup$ @DietrichBurde What do you mean? k^2 = 1 + 2*2^2 + ... + l * l^2 This exact line appears in the right hand side doesn't it? $\endgroup$
    – user176799
    Sep 17 '14 at 19:13
  • $\begingroup$ @Darksonn Thank you very much, I'll try to familiarize myself with it $\endgroup$
    – user176799
    Sep 17 '14 at 19:15
  • $\begingroup$ @DietrichBurde But can't I plug from the first equation into the second one? If k^2 = 1 + 2*2^2 + ... + l * l^2, and 1 + 2*2^2 + ... + l * l^2 appears in the second last equation, can't I substitute the whole thing for k^2? That was my understanding on how induction worked... $\endgroup$
    – user176799
    Sep 17 '14 at 19:20
1
$\begingroup$

You want to show that $$ k(n)^2=1\cdot1^2+2\cdot2^2+3\cdot3^2+\ldots+n\cdot n^2, $$ where $k(n)=1+2+\cdots+n$, for any $n$. For $n=1$, you have $k(1)^2=1^2=1$ on the left and $1\cdot 1^2=1$ on the right. Now assuming the equation holds for $n$, you want to show that it also holds for $n+1$. Note that $k(n+1)=k(n)+n+1$. Now $$ k(n+1)^2=\left(k(n)+n+1\right)^2=k(n)^2+(n+1)\left(2 k(n)+n+1\right)\\=1\cdot 1^2 + 2\cdot 2^2 + \ldots + n\cdot n^2+(n+1)\left(2 k(n)+n+1\right). $$ For this to help you, you need to also show that $2k(n)+n+1=(n+1)^2$, or that $k(n)=\frac{1}{2}(n+1)^2-\frac{1}{2}(n+1)=\frac{1}{2}n(n+1).$ (You can show that by induction, too, if you like.) Once that step is done, you'll have shown that the equation holds for $n=1$, and that if it holds for some $n$ it also holds for $n+1$; and you can conclude by induction that it holds for all $n\ge 1$.

$\endgroup$
2
  • $\begingroup$ Sorry, I don't understand this very well... Would you mind clarifying it a bit more? First thing: why do I want to show $$ k(n)^2=1\cdot1^2+2\cdot2^2+3\cdot3^2+\ldots+n\cdot n^2? $$ Is what I'm trying to prove not $$ k^2=1\cdot1^2+2\cdot2^2+3\cdot3^2+\ldots+n\cdot n^2 $$ Why $k(n)^2$ instead of $k^2$? $\endgroup$
    – user176799
    Sep 17 '14 at 19:33
  • $\begingroup$ I wrote $k(n)$ because $k$ is a function of $n$, defined by $k(n)=1+2+\cdots+n$. Just writing "$k$" without showing the dependence on $n$, especially when more than one value of $n$ will be considered, can be confusing. $\endgroup$
    – mjqxxxx
    Sep 18 '14 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.